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scottman
How do I find the complex cube root of 8?
method one solve \[x^3=8\] \[x^3-8=0\] \[(x-2)(x^2+2x+4)=0\] quadratic formula will give the solutions to \[x^2+2x+4=0\]
I see and so 2 is one root and the other 2 complex numbers will be the other 2 roots?
method two solve \(x^3=1\) and multiply by two since one solution is 1, divide the unit circle in two three equal parts one is at 1, the other two are \(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\) and its conjugate \(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\) multiply these numbers by 2
to answer your question, yes, the solution to the quadratic will give you the two complex roots
ok thanks a lot, you helped me greatly